In the previous section we completed a discussion on the basics of 'electric power'. We also saw some solved examples. In this section, we will see a few more solved examples.
Solved example 10.8
A bulb of resistance 529 Ω works on 230 V. The filament was broken. It was joined together and made to work. Let the resistance of the filament now be 460 Ω.
(a) What is the change in power of the bulb?
(b) What is the current flowing through the bulb?
Solution:
Part (a):
1. Given: R1 = 529 Ω, R2 = 460 Ω, V = 230 volts.
2. When the bulb was manufactured, it had a certain resistance. It was 529 Ω.
• So the manufacturer knows how much energy it will consume in one second (if 230 V is supplied) .
• We can calculate it ourselves using the equation which relates power to [voltage, resistance]
• We can use Eq.10.5: P = V2⁄R
• Substituting the values, we get: P1 = 2302⁄529 = 100 W.
3. When the new resistance is 460, the new power = P2 = 2302⁄460 = 115 W.
So change in power = (115 - 100) = 15 W
4. The power has increased. That means, more energy is consumed, which is equivalent to saying: 'more energy is produced'.
• So more heat will be produced. But the filament may not be able to withstand this greater heat. Thus it will melt and break again
5. Another explanation:
• We have seen in the previous section that, when the broken pieces of the filament are joined together, the current will increase.
• According to Joule's law, when current increases, heat produced also increases
• But the filament may not be able to withstand this greater heat. Thus it will melt and break again
Part (b):
1. We are required to find the currents.
We can use Eq.10.6: P = VI, which relates power to [Voltage, current]
2. Voltage remains constant at 230 V
So we can write:
• P1 = V × I1 ⟹100 = 230 × I1 ⟹ I1 = 100⁄230 = 0.4348 A
• P2 = V × I2 ⟹115 = 230 × I2 ⟹ I2 = 115⁄230 = 0.5 A
3. So change in current = (I2 - I1) = (0.5 - 0.4348) = 0.0652 A
Solved example 10.9
A bulb is designed to work in 230 V. The resistance of the filament is 529 Ω. If the bulb works at 115 V, what will be it's power?
Solution:
1. Given: V1 = 230 V, V2 = 115 V, R = 529 Ω.
2. The manufacturer designed the bulb to work at 230 V. For that, a resistance of 529 Ω was given
• We can use Eq.10.5: P = V2⁄R, which relates power to [Voltage, Resistance]
3. We can write: P2 = (V2)2⁄R
• Substituting the values, we get: P2 = (115)2⁄529 = 25 W
■ Initial power = P1 = (230)2⁄529 = 100 W
Note that, there is no change in resistance. In such a situation, if voltage is halved (ie., from 230 to 115V), the power becomes one fourth (ie., from 100 to 25)
Solved example 10.10
Two bulbs are marked as follows:
(i) 40 W, 240 V
(ii) 100 W, 240 V
Which one has greater resistance?
Solution:
■ We want an equation which relates power to: [voltage, resistance]
• We can use Eq.10.5: P = V2⁄R,
Case (i): Substituting the values, we get: 40 = 2402⁄R ⟹ R = 2402⁄40 = 1440 Ω
Case (ii): Substituting the values, we get: 100 = 2402⁄R ⟹ R = 2402⁄100 = 576 Ω
Solved example 10.11
An electric heater of power 920 W is working on a 230 V supply. If current flows for 5 minutes through it, Calculate the heat generated
Solution:
1. From Joule's law, we have: Heat generated (H) = I2Rt
2. So we want 3 items:
• Current flowing through the heater (I)
• Resistance of the heater (R)
• Time for which current flows through the heater (t)
3. In the question, we are given only 't'
• We have to calculate the other two items ourselves
4. To calculate them, we can use the given P and V
• To find I, we want a relation that will connect the power (P) to: [V, I]
• We will use Eq.10.6: P = VI
• Substituting the values, we get: 920 = 230 × I ⟹ I = 920⁄230 = 4 A
5. To find R, we want a relation that will connect the power (P) to: [V, I]
• We will use Eq.10.5: P = V2⁄R
• Substituting the values, we get: 920 = 2302⁄R ⟹ R = 2302⁄920 = 57.5 Ω
6. Now we have the required items. Substituting the values in (1),we get:
H = I2Rt = 42 × 57.5 × (5 × 60) = 276000 joules
7. Another method:
• After calculating I in step (4), we can directly use Eq.10.3: H = VIt (Details here)
• In this method, we need not calculate R
• Substituting the values, we get: H = 230 × 4 × (5 × 60) = 276000 joules
Solved example 10.12
The marking on an electrical appliance is 800 W, 200 V
(a) If it works on 100 V, what is the power consumed?
(b) What is the power when it works on 50 V?
(c) What happens if 500 V is applied to it?
Solution:
• The marked values are: 800 W and 200 V
• The manufacturer is telling us that, if we supply 200 V to that appliance, it will do a work of 800 joules every second
• Another way of saying it is:
If we supply 200 V to that appliance, it will consume 800 joules of energy every second
• Yet another way of saying it is:
If we supply 200 V to that appliance, it will consume a power of 800 W
• The appliance has a fixed resistance R. We have to find this R first.
• So we want a relation that will connect the power (P) to: [R, V]
• We will use Eq.10.5: P = V2⁄R
3. Substituting the values, we get: 800 = 2002⁄R ⟹ R = 2002⁄800 = 50 Ω.
Part (a):
1. If the voltage applied is 100 V (instead of 200 V), we will not get the 800 W power. It will be less
2. We have to find this new power. So we want a relation that will connect the power (P) to: [R, V]
• We will use Eq.10.5: P = V2⁄R
3. Substituting the values, we get: P = 1002⁄50 = 200 W
■ Note:
• The voltage decreased from 200 to 100 V. That is., new voltage is half of original
♦ Then the power decreased from 800 to 200 W. That is., power becomes one fourth of the original
Part (b):
1. If the voltage applied is 50 V (instead of 200 V), we will not get the 800 W power. It will be less
2. We have to find this new power. So we want a relation that will connect the power (P) to: [R, V]
• We will use Eq.10.5: P = V2⁄R
3. Substituting the values, we get: P = 502⁄50 = 50 W
■ Note:
• The voltage decreased from 200 to 50 V. That is., new voltage is one fourth of original
♦ Then the power decreased from 800 to 50 W.
♦ But 50 = (800 × 1⁄16)
♦ That is., power becomes one sixteenth of the original
Part (c):
1. 500 V is greater than the marked value of 200 V.
• A greater voltage will cause damage to the appliance. Let us see how:
2. From Ohm's law, we have: V = IR
• Substituting the values, we get: 500 = I × 50 ⟹ I = 10 A
3. Using the same method, we can calculate the current expected by the manufacturer:
V = IR ⟹ 200 = I × 50 ⟹ I = 4 A
4. So the manufacturer is expecting a current of only 4 A
• We would be supplying 10 A
• The conductor in that appliance will not be able to carry 10 A. It will melt, and thus the appliance will be damaged.
Solved example 10.13
A potential difference of 200 V is applied to a 200 Ω resistor for 5 minutes
(a) Calculate the amount of heat generated
(b) If 200 Ω is replaced by 100 Ω, how much heat will be generated in 5 minutes?
(c) If 400 Ω resistance is used in this place for 5 minutes, how much heat is generated?
Solution:
Part (a):
1. From Joule's law, we have: Heat generated (H) = I2Rt
2. So we want 3 items:
• Current flowing through the resistor (I)
• Resistance (R)
• Time for which current flows through the resistor (t)
3. In the question, we are given 'R' and 't'
• We have to calculate 'I' ourselves
4. From Ohm's law, we have: V = IR
• Substituting the values, we get: 200 = I × 200 ⟹ I = 1 A
5. So now we have all the items. Substituting in (1), we get:
H = I2Rt = 12 × 200 × (5 × 60) = 60000 joules
Part (b):
1. From Joule's law, we have: Heat generated (H) = I2Rt
2. So we want 3 items:
• Current flowing through the resistor (I)
• Resistance (R)
• Time for which current flows through the resistor (t)
3. In the question, we are given 'R' and 't'
• We have to calculate 'I' ourselves
4. From Ohm's law, we have: V = IR
Substituting the values, we get: 200 = I × 100 ⟹ I = 2 A
5. So now we have all the items. Substituting in (1), we get:
H = I2Rt = 22 × 100 × (5 × 60) = 120000 joules
Part (c):
1. From Joule's law, we have: Heat generated (H) = I2Rt
2. So we want 3 items:
• Current flowing through the resistor (I)
• Resistance (R)
• Time for which current flows through the resistor (t)
3. In the question, we are given 'R' and 't'
• We have to calculate 'I' ourselves
4. From Ohm's law, we have: V = IR
Substituting the values, we get: 200 = I × 400 ⟹ I = 0.5 A
5. So now we have all the items. Substituting in (1), we get:
H = I2Rt = (0.5)2 × 400 × (5 × 60) = 30000 joules
Solved example 10.14
Correct the wrong equations:
H = I2Rt, H = V2Rt, H = Pt, H = IVt
Solution:
• 'H = V2Rt' is wrong.
• The correct form is: H = I2Rt
We have completed this discussion on 'effects of electric current' and also 'electric power'.
• In a heater, electrical energy is converted into heat energy
• In an incandescent bulb, electrical energy is converted into both heat energy and light energy
• In an electric motor, electrical energy is converted into mechanical energy
■ A great advantage of electrical energy is that, it can be easily converted into other forms of energy
• But we must use appropriate appliances for the conversion.
Solved example 10.8
A bulb of resistance 529 Ω works on 230 V. The filament was broken. It was joined together and made to work. Let the resistance of the filament now be 460 Ω.
(a) What is the change in power of the bulb?
(b) What is the current flowing through the bulb?
Solution:
Part (a):
1. Given: R1 = 529 Ω, R2 = 460 Ω, V = 230 volts.
2. When the bulb was manufactured, it had a certain resistance. It was 529 Ω.
• So the manufacturer knows how much energy it will consume in one second (if 230 V is supplied) .
• We can calculate it ourselves using the equation which relates power to [voltage, resistance]
• We can use Eq.10.5: P = V2⁄R
• Substituting the values, we get: P1 = 2302⁄529 = 100 W.
3. When the new resistance is 460, the new power = P2 = 2302⁄460 = 115 W.
So change in power = (115 - 100) = 15 W
4. The power has increased. That means, more energy is consumed, which is equivalent to saying: 'more energy is produced'.
• So more heat will be produced. But the filament may not be able to withstand this greater heat. Thus it will melt and break again
5. Another explanation:
• We have seen in the previous section that, when the broken pieces of the filament are joined together, the current will increase.
• According to Joule's law, when current increases, heat produced also increases
• But the filament may not be able to withstand this greater heat. Thus it will melt and break again
Part (b):
1. We are required to find the currents.
We can use Eq.10.6: P = VI, which relates power to [Voltage, current]
2. Voltage remains constant at 230 V
So we can write:
• P1 = V × I1 ⟹100 = 230 × I1 ⟹ I1 = 100⁄230 = 0.4348 A
• P2 = V × I2 ⟹115 = 230 × I2 ⟹ I2 = 115⁄230 = 0.5 A
3. So change in current = (I2 - I1) = (0.5 - 0.4348) = 0.0652 A
Solved example 10.9
A bulb is designed to work in 230 V. The resistance of the filament is 529 Ω. If the bulb works at 115 V, what will be it's power?
Solution:
1. Given: V1 = 230 V, V2 = 115 V, R = 529 Ω.
2. The manufacturer designed the bulb to work at 230 V. For that, a resistance of 529 Ω was given
• We can use Eq.10.5: P = V2⁄R, which relates power to [Voltage, Resistance]
3. We can write: P2 = (V2)2⁄R
• Substituting the values, we get: P2 = (115)2⁄529 = 25 W
■ Initial power = P1 = (230)2⁄529 = 100 W
Note that, there is no change in resistance. In such a situation, if voltage is halved (ie., from 230 to 115V), the power becomes one fourth (ie., from 100 to 25)
Solved example 10.10
Two bulbs are marked as follows:
(i) 40 W, 240 V
(ii) 100 W, 240 V
Which one has greater resistance?
Solution:
■ We want an equation which relates power to: [voltage, resistance]
• We can use Eq.10.5: P = V2⁄R,
Case (i): Substituting the values, we get: 40 = 2402⁄R ⟹ R = 2402⁄40 = 1440 Ω
Case (ii): Substituting the values, we get: 100 = 2402⁄R ⟹ R = 2402⁄100 = 576 Ω
Solved example 10.11
An electric heater of power 920 W is working on a 230 V supply. If current flows for 5 minutes through it, Calculate the heat generated
Solution:
1. From Joule's law, we have: Heat generated (H) = I2Rt
2. So we want 3 items:
• Current flowing through the heater (I)
• Resistance of the heater (R)
• Time for which current flows through the heater (t)
3. In the question, we are given only 't'
• We have to calculate the other two items ourselves
4. To calculate them, we can use the given P and V
• To find I, we want a relation that will connect the power (P) to: [V, I]
• We will use Eq.10.6: P = VI
• Substituting the values, we get: 920 = 230 × I ⟹ I = 920⁄230 = 4 A
5. To find R, we want a relation that will connect the power (P) to: [V, I]
• We will use Eq.10.5: P = V2⁄R
• Substituting the values, we get: 920 = 2302⁄R ⟹ R = 2302⁄920 = 57.5 Ω
6. Now we have the required items. Substituting the values in (1),we get:
H = I2Rt = 42 × 57.5 × (5 × 60) = 276000 joules
7. Another method:
• After calculating I in step (4), we can directly use Eq.10.3: H = VIt (Details here)
• In this method, we need not calculate R
• Substituting the values, we get: H = 230 × 4 × (5 × 60) = 276000 joules
Solved example 10.12
The marking on an electrical appliance is 800 W, 200 V
(a) If it works on 100 V, what is the power consumed?
(b) What is the power when it works on 50 V?
(c) What happens if 500 V is applied to it?
Solution:
• The marked values are: 800 W and 200 V
• The manufacturer is telling us that, if we supply 200 V to that appliance, it will do a work of 800 joules every second
• Another way of saying it is:
If we supply 200 V to that appliance, it will consume 800 joules of energy every second
• Yet another way of saying it is:
If we supply 200 V to that appliance, it will consume a power of 800 W
• The appliance has a fixed resistance R. We have to find this R first.
• So we want a relation that will connect the power (P) to: [R, V]
• We will use Eq.10.5: P = V2⁄R
3. Substituting the values, we get: 800 = 2002⁄R ⟹ R = 2002⁄800 = 50 Ω.
Part (a):
1. If the voltage applied is 100 V (instead of 200 V), we will not get the 800 W power. It will be less
2. We have to find this new power. So we want a relation that will connect the power (P) to: [R, V]
• We will use Eq.10.5: P = V2⁄R
3. Substituting the values, we get: P = 1002⁄50 = 200 W
■ Note:
• The voltage decreased from 200 to 100 V. That is., new voltage is half of original
♦ Then the power decreased from 800 to 200 W. That is., power becomes one fourth of the original
Part (b):
1. If the voltage applied is 50 V (instead of 200 V), we will not get the 800 W power. It will be less
2. We have to find this new power. So we want a relation that will connect the power (P) to: [R, V]
• We will use Eq.10.5: P = V2⁄R
3. Substituting the values, we get: P = 502⁄50 = 50 W
■ Note:
• The voltage decreased from 200 to 50 V. That is., new voltage is one fourth of original
♦ Then the power decreased from 800 to 50 W.
♦ But 50 = (800 × 1⁄16)
♦ That is., power becomes one sixteenth of the original
Part (c):
1. 500 V is greater than the marked value of 200 V.
• A greater voltage will cause damage to the appliance. Let us see how:
2. From Ohm's law, we have: V = IR
• Substituting the values, we get: 500 = I × 50 ⟹ I = 10 A
3. Using the same method, we can calculate the current expected by the manufacturer:
V = IR ⟹ 200 = I × 50 ⟹ I = 4 A
4. So the manufacturer is expecting a current of only 4 A
• We would be supplying 10 A
• The conductor in that appliance will not be able to carry 10 A. It will melt, and thus the appliance will be damaged.
Solved example 10.13
A potential difference of 200 V is applied to a 200 Ω resistor for 5 minutes
(a) Calculate the amount of heat generated
(b) If 200 Ω is replaced by 100 Ω, how much heat will be generated in 5 minutes?
(c) If 400 Ω resistance is used in this place for 5 minutes, how much heat is generated?
Solution:
Part (a):
1. From Joule's law, we have: Heat generated (H) = I2Rt
2. So we want 3 items:
• Current flowing through the resistor (I)
• Resistance (R)
• Time for which current flows through the resistor (t)
3. In the question, we are given 'R' and 't'
• We have to calculate 'I' ourselves
4. From Ohm's law, we have: V = IR
• Substituting the values, we get: 200 = I × 200 ⟹ I = 1 A
5. So now we have all the items. Substituting in (1), we get:
H = I2Rt = 12 × 200 × (5 × 60) = 60000 joules
Part (b):
1. From Joule's law, we have: Heat generated (H) = I2Rt
2. So we want 3 items:
• Current flowing through the resistor (I)
• Resistance (R)
• Time for which current flows through the resistor (t)
3. In the question, we are given 'R' and 't'
• We have to calculate 'I' ourselves
4. From Ohm's law, we have: V = IR
Substituting the values, we get: 200 = I × 100 ⟹ I = 2 A
5. So now we have all the items. Substituting in (1), we get:
H = I2Rt = 22 × 100 × (5 × 60) = 120000 joules
Part (c):
1. From Joule's law, we have: Heat generated (H) = I2Rt
2. So we want 3 items:
• Current flowing through the resistor (I)
• Resistance (R)
• Time for which current flows through the resistor (t)
3. In the question, we are given 'R' and 't'
• We have to calculate 'I' ourselves
4. From Ohm's law, we have: V = IR
Substituting the values, we get: 200 = I × 400 ⟹ I = 0.5 A
5. So now we have all the items. Substituting in (1), we get:
H = I2Rt = (0.5)2 × 400 × (5 × 60) = 30000 joules
Solved example 10.14
Correct the wrong equations:
H = I2Rt, H = V2Rt, H = Pt, H = IVt
Solution:
• 'H = V2Rt' is wrong.
• The correct form is: H = I2Rt
We have completed this discussion on 'effects of electric current' and also 'electric power'.
• In a heater, electrical energy is converted into heat energy
• In an incandescent bulb, electrical energy is converted into both heat energy and light energy
• In an electric motor, electrical energy is converted into mechanical energy
■ A great advantage of electrical energy is that, it can be easily converted into other forms of energy
• But we must use appropriate appliances for the conversion.
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